我正在尝试在波纹管表上编写一个 sql 查询。
╔════╦══════════╦═══════╗======╗======╗
║ ID ║ NAME ║ CLASS ║PARENT║ DOB ║
╠════╬══════════╬═══════╣======║======║
║ 1 ║ DAVID ║ SPIN ║ ║1 ║
║ 2 ║ AROON ║ BIKE ║ 1 ║1 ║
║ 3 ║ LEO ║ YOGA ║ ║2 ║
║ 4 ║ LIN ║ CYC ║ 1 ║2 ║
║ 5 ║ STEFA ║ YOGA ║ ║3 ║
║ 6 ║ GLORIA ║ RUNN ║ 1 ║3 ║
╚════╩══════════╩═══════╝======╝======╝
并且,此表的输出应如下所示
╔════╦════════╦═══════╗======╗======╗
║ ID ║ NAME ║ CLASS ║PARENT║ DOB ║
╠════╬════════╬═══════╣======║======║
║ 1 ║ DAVID ║ SPIN ║ ║1 ║
║ 2 ║ AROON ║ BIKE ║ 1 ║1 ║
║ 4 ║ LIN ║ CYC ║ 1 ║2 ║
║ 6 ║ GLORIA║ RUNN ║ 1 ║3 ║
║ 3 ║ LEO ║ YOGA ║ ║2 ║
║ 5 ║ STEFAN║ YOGA ║ ║3 ║
╚════╩════════╩═══════╝======╝======╝
So this is the explanation of the output
First parent David as his DOB is 1,
--David three childrens sorted based on DOB
Then LEO as his DOB is 2
-- Leo do not have children[if he did, would be here as sorted on DOB]
Then Stefan as his DOB is 3
-- Stefan do not have children [if he did, would be here as sorted on DOB]
那么我尝试了什么?
SELECT * FROM user group by ID, PARENT ;
在 SQL 之上,语句返回父子组中的项目但不保持任何顺序,当我添加 ORDER BY
时,SQL
似乎不尊重 GROUP BY了。
然后我尝试连接并以两个完全不同的表结束,其中一个包含所有 parent ,另一个包含所有 child 。 UNION ALL
对这两个查询返回了预期的数据集,但未按预期顺序返回。
有什么想法吗?
更新
Output should be
Pick entry [based on min time ].
--use that id and find all of its children and placed them in sorted order
repeat for every row in the table
注意:
--parents are sorted based on DOB
--child's are also sorted based on DOB
--DOB are valid timestamp
--PARENT, ID field both are UUID and define as CHAR, PARENT reference to ID
更新 1
下面的查询
WITH RECURSIVE
top AS (
SELECT * FROM (SELECT * FROM user WHERE PARENT is null ORDER BY dob LIMIT 1)
UNION
SELECT user.NAME, user.PARENT, user.ID, user.CLASS, user.DOB FROM user, top WHERE user.PARENT=top.ID
ORDER BY user.dob
) SELECT * FROM top;
返回以下输出:
╔════╦════════╦═══════╗======╗======╗
║ ID ║ NAME ║ CLASS ║PARENT║ DOB ║
╠════╬════════╬═══════╣======║======║
║ 1 ║ DAVID ║ SPIN ║ ║1 ║
║ 2 ║ AROON ║ BIKE ║ 1 ║1 ║
║ 4 ║ LIN ║ CYC ║ 1 ║2 ║
║ 5 ║ GLORIA║ RUNN ║ 1 ║3 ║
╚════╩════════╩═══════╝======╝======╝
输出对第一个 parent 有利。但是,仍然无法弄清楚,我怎样才能按排序顺序遍历其余的 parent 和他们的 child 。
最佳答案
查询
SELECT u1.*
FROM `user` u1
LEFT JOIN `user` u2
ON u1.PARENT = u2.ID
ORDER BY CASE WHEN u1.PARENT IS NULL THEN u1.DOB ELSE u2.DOB END
|| CASE WHEN u1.PARENT IS NULL THEN '' ELSE u1.DOB END;
解释
- 别名
u1
包含所有用户详细信息 - Alias
u2
包含父级的详细信息(如适用)。 (使用了LEFT JOIN
,因此如果u1
用户没有父级,这些详细信息将全部为null
。) - 如果用户没有 parent ,则单独使用其 DOB 进行订购。
- 如果用户有 parent ,则获取用户 parent 的 DOB 并连接(追加)用户( child )的 DOB。
结果
用于 ORDER BY
的构造值(在 SELECT
中实际上不需要)看起来像这里最右边的列:
╔════╦════════╦═══════╗======╗======╦════════╗
║ ID ║ NAME ║ CLASS ║PARENT║ DOB ║ORDER BY║
╠════╬════════╬═══════╣======║======╬════════║
║ 1 ║ DAVID ║ SPIN ║ ║1 ║ 1 ║
║ 2 ║ AROON ║ BIKE ║ 1 ║1 ║ 11 ║
║ 4 ║ LIN ║ CYC ║ 1 ║2 ║ 12 ║
║ 6 ║ GLORIA║ RUNN ║ 1 ║3 ║ 13 ║
║ 3 ║ LEO ║ YOGA ║ ║2 ║ 2 ║
║ 5 ║ STEFAN║ YOGA ║ ║3 ║ 3 ║
╚════╩════════╩═══════╝======╝======╩════════╝
演示
参见 SQL Fiddle Demo .
关于android - 父子关系的SQL查询,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/35736610/